Equally Likely Outcomes
Roll a fair die. Flip a fair coin. Draw a card, face-down, from a well-shuffled pack. In each of
these there is something special going on: every outcome has exactly the same chance
as every other. No face of the die is favoured; heads is no likelier than tails; the two of clubs is
no easier to pull out than the queen of hearts.
When that is true, something wonderful happens — you don't need to flip, roll, or draw
anything to find a probability. You just count. Count how many outcomes make
your event happen, count how many outcomes there are altogether, and divide.
P(\text{event}) = \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}
A fair coin has two equally likely faces, one of which is heads, so
P(\text{heads}) = \tfrac{1}{2}. A fair die has six equally likely faces,
and three of them are even (2, 4, 6), so
P(\text{even}) = \tfrac{3}{6} = \tfrac{1}{2}. Counting, not guessing.
When every outcome is equally likely:
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the probability of an event is
P = \dfrac{\text{favourable}}{\text{total}};
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so the whole job is to count carefully — how many outcomes are favourable, and
how many there are altogether;
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a fair die has 6 equally likely faces, a coin has 2, and a single draw from a bag has one
outcome per object in the bag.
Every probability lives between 0 and 1
Because the favourable outcomes are always part of the total, the top of that fraction can
never be bigger than the bottom. So every probability sits somewhere on a line from
0 to 1:
- 0 means impossible — no favourable outcomes at all
(P(\text{rolling a 7 on a die}) = \tfrac{0}{6} = 0).
- 1 means certain — every outcome is favourable
(P(\text{rolling less than 7}) = \tfrac{6}{6} = 1).
- Everything else — a coin's \tfrac12, a die's
\tfrac16 — lands in between.
If you ever compute a probability bigger than 1 or below 0, stop: something has gone wrong in the
counting.
Worked example 1 — an even number on a die
Lay out the six faces of a fair die, then highlight the ones that count as "even". Three out of six
— that is the probability.
There are 6 equally likely faces. The favourable ones (even) are
2, 4, 6 — that's 3 of them. So
P(\text{even}) = \frac{3}{6} = \frac{1}{2} = 0.5.
Notice it doesn't matter which even face turns up — we only counted how many.
Worked example 2 — a red card from a pack
A standard pack has 52 cards, all equally likely to be the one you draw.
Exactly half of them — the hearts and the diamonds — are red, so
there are 26 favourable cards:
P(\text{red}) = \frac{26}{52} = \frac{1}{2} = 0.5.
The same counting handles trickier events. There are 13 hearts, so
P(\text{heart}) = \tfrac{13}{52} = \tfrac14 = 0.25. There are only
4 aces, so
P(\text{ace}) = \tfrac{4}{52} = \tfrac{1}{13} \approx 0.08. Big total, small
total — the recipe never changes: favourable over total.
Worked example 3 — a spinner that breaks the rule
Here is a spinner with three colours: red, blue and green. It is tempting to say "three outcomes, so
P(\text{red}) = \tfrac13." Wrong. Look at the sizes: red
takes up half the circle, while blue and green each take a quarter.
The three colours are not equally likely, so you cannot just count colours. Here the honest
probabilities follow the areas: P(\text{red}) = \tfrac12,
P(\text{blue}) = \tfrac14,
P(\text{green}) = \tfrac14. The counting shortcut only works when the pieces
being counted are genuinely the same size.
Two traps catch almost everyone:
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The counting formula only works when the outcomes are genuinely equally likely. A
bent coin, a weighted die, or a spinner with different-sized sectors are
not equally likely — so you can't simply count. Counting "three colours" on the
unfair spinner gives the wrong answer. When symmetry is broken, you must
experiment
instead.
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Probability is favourable ÷ TOTAL — never favourable ÷ unfavourable. For an even
number on a die there are 3 even faces and 3 odd faces, but
P(\text{even}) is \tfrac{3}{6}, not
\tfrac{3}{3}. The bottom of the fraction is everything that
could happen, favourable outcomes included.
Yes — and quite a famous one. In the 1650s a French gambler, the Chevalier de Méré, was stuck on a
puzzle: if a dice game has to stop early, how should the players fairly split the stakes,
given who was ahead? He put it to the mathematician Blaise
Pascal, who wrote to Pierre de Fermat,
and the two swapped letters working out the answer by carefully counting the equally likely ways
the unfinished game could have ended.
Out of that argument about dice money grew the whole subject of probability theory —
the maths behind weather forecasts, insurance, medicine, and every game you'll ever play. Not bad for
a quarrel over a bet.
See it explained